Question

In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $I$, $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda }{4}$ will be

• A. $\frac{I}{4}$
• B. $\frac{1}{2}$
• C. $I$
• D. zero

Answer 1

Answer: (B)

For path difference $\lambda$, phase
difference $=2\pi \left(Q=\frac{2\pi }{\lambda }x=\frac{2\pi }{\lambda }\cdot \lambda =2\pi \right)$
$\Rightarrow I=I_0+I_0+2I_0\cos 2\pi$
$\Rightarrow I=4I_0$
For $x=\frac{\lambda }{4}$, phase difference $=\frac{\pi }{2}$
$\therefore I^{\prime }=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \frac{\pi }{2}$
If $I_1=I_2=I_0$ then $I^{\prime }=2I_0=2\cdot \frac{I}{4}=\frac{I}{2}$